Proof for Trignometrical ratios of 90 + theta

[milan.rajik]

Please somebody give me proofs for Trignometrical ratios of \[\sin\](90 + \[\theta\]) and \[\sin\](180 - \[\theta\]). I need geometrical proof.

 

[albbg]

If you see at the unit circle:

1. rotating by 90 degrees means swapping x and y axis (that is sin and cos), then sin(90+θ)=cos(θ)
2. rotating 180 degrees the waveform repeats with reversed sign, that means sin(180-θ)=-sin(-θ)=sin(θ)

then the ratio sin(90+θ)/sin(180-θ) = cotg(θ)

You can see the same using the trigonometric equality:

sin(x+y)=sin(x)*cos+cos(x)*sin then:

sin(90+θ)=sin(90)*cos(θ)+cos(90)*sin(θ)=cos(θ)
sin(180-θ)=sin(180)*cos(-θ)+cos(180)*sin(-θ)=-sin(-θ)=sin(θ)

 

[milan.rajik]

I don't need unit circle proof or proof using trignometric equality. I know those. I am attaching a image. Please read it. It is from the book Plane Trigonometry by S L Loney. See what it says.
I know that angle MOP = angle M'P'O and both triangles formed are congruent, OM' = MP and M'P" = OM and OP' = OP.

Angle P'OM' + angle MOP = 90.
If angle MOP is theta then angle M'P'O is also theta.
Angle OPM = angle P'OM' = 90 - theta because 3 angles of triangle = 180. ( 90 + theta + (90 - theta))

What I don't understand is how is andle MOP = 90 - angle P'OM' and how it is = to angle OP'M'. ( as OP'M' = theta)

 

[rahdirs]

What I don't understand is how is angle MOP = 90 - angle P'OM' and how it is = to angle OP'M'. ( as OP'M' = theta)

You must be knowing MOM' is a straight line.
Thus, ∠MOP + ∠POP' + ∠P'OM' = 180,
we already know that ∠POP' = 90.
Hence ∠MOP = 180 - 90 - ∠P'OM'= 90 - ∠P'OM'.

In triangle P'OM',
Sum of three angles = ∠P'OM' + ∠OP'M' + ∠P'M'O = 180.
we already know that ∠P'M'O = 90
thus ∠P'OM' + ∠OP'M' = 90 & ∠P'OM' = 90- ∠MOP(from above, ∠MOP = 90- ∠P'OM')
Thus, ∠OP'M' = ∠MOP.

Are you sure,this is what you want ?????

 

[milan.rajik]

Yes. I understand that, but what is confusing me is how can a angle have two different values. Let me explain.

angle POP' = 90
angle MOP + p'OM' = 90
In triangle MOP, angle MOP = theta (say), the other two angles are 90 and (90 - theta)
In triangle P'OM', angle M'P'O = theta, and angle P'OM' is (90 - theta) and another angle is 90.

Let us put angle angle P'OM' as angle A and angle MOP as B then A + B = 90, and B = 90 - A. (A is already 90 - theta). So, B = 90 - (90 - theta) = theta.

Where are we taking sin(90 + theta). we are just taking sin(theta) in second quadrant. where did the 90 in (sin(90 + theta) go?

 

[rahdirs]

• there seems to be some error in first three lines.
• Initially he started from OA & trace out an angle theta in clock-wise direction & reached OP.Hence ∠AOP = theta.(agreed with book)
• Now he revolved it in 90 positive direction to reach OP',so ∠AOP' = 90 + theta !!!!!!!!!!!!(it should've been 90-theta).

Now as to your question,where do you have an error,at the line sin(90+theta) = sin(AOP') ?????????//

--------------------Updated-----------------------------
He did some error,over there.Initially in the third line,he took ∠AOP = theta & at line 15,where he wrote Sin(90 + theta) = Sin(∠AOP'),∠AOP' = 90+theta & ∠AOP = theta & ∠POP' = 90!!!!!!!!!!
Clearly some misprint.

 

[milan.rajik]

The point A in the diagram is wrong. That is my mistake. Point A is to the left of M on x-axis.

There is no error. angle MOP or AOP is theta. AOP' and MOP' are both same. Point A is just to the right of M and A' to the left of M'. My question is how did he calculate
sin(90 + theta) = sin(AOP') = M'P'/OP' = OM/OP. This is giving a problem. M'P'/OP' is cos(theta) in second quardrant.
If we take angle MOP as theta and angle P'OM' as (90 - theta) as sum of the angles should be 90. then angle P'OM' is 90 - theta.
Again if we consider just the triangle in 2nd quadrant then the angle M'P'O is theta and angle P'M'O is 90 and hence the other angle of the triangle is (90 - theta).

M'P'/OP' is sin(90 - theta) of triangle in 2nd quadrant or cos(theta) of triangle in first quadrant.

Then how does he say that M'P'/OP' is sin(90 + theta)?

So many people refer the book by S L Loney. I am attaching a new image.

 

[albbg]

I think rahdirs explained all in his post #4. You have two triangles with the same hypotenuse (OP=OP') and three congruent angles, that means the catheti associated to each angle must be the same between the two figures. This means P will be associated to the same catheti of O', M to the same of M' and O to the same of P'. Thus M'P'/OP'=OM/OP

 

[rahdirs]

This means P will be associated to the same catheti of O', M to the same of M' and O to the same of P'. Thus M'P'/OP'=OM/OP

@albbg:Yes,it was shown in post#4 that M'P'/OP' = OM/OP.But @milan wanted to know how it was written that sin(∠AOP') = M'P'/OP' ????If that is done,then it can be shown that M'P'/OP' = OM/OP & then,sin(90+theta) = sin(∠AOP') = OM/OP = cos(theta).

Then how does he say that M'P'/OP' is sin(90 + theta)?

You should've posted all four images.
In fig. 1 & fig.2 as in post #4,∠AOP = theta,∠AOP' = 90 + theta. Sin(90 + theta) = sin(∠AOP') = (M'P'/OP') (but how ????)

He seems to have taken that theta could be arbitrary either in 1st/2nd/3rd/4th quadrant as in four diagrams attached.
From help of fig.3 & fig.4 (bottom two),you could easily see that sin(∠AOP') = (M'P'/OP').

But sin(∠AOP') = (M'P'/OP') should be valid only in those bottom two cases & shouldn't be generalized.I need to think it a bit on how he extended it to above two diagrams.

 

[jayanth.devarayanadurga]

Did you make a mistake. Where is MP'/OP'. Was it M'P'/OP'.

 

[rahdirs]

Yes,wrote M'P' as MP'.Changed all MP' to M'P'.

 

[milan.rajik]

Yes. In fig 4. P'M'/OP' is sin(theta) but not sin(90 + theta).

 

[rahdirs]

@milan: In fig.3 & fig.4(bottom two) M'P'/OP' = Sin(∠AOP') & it is not Sin(theta),it is Sin(270-theta).

Because theta in those two cases ∠AOP(theta) is obtuse angle & ∠AOP' is 90 + theta (obtuse one,because OP' has been brought by rotating OP by 90 & OP is brought by rotating OA by theta)because ∠POP' is 90.

But in that triangle, M'P'/OP' = Sin(∠AOP').But ∠AOP' in that triangle,in fig.3 is 270 - theta & in fig.4 is theta - 270.

I somehow get the feeling that he extended the fact that Sin(∠AOP') = M'P'/OP' from fig.3 & fig.4 to irrespective of what ∠AOP' might be.

 

[jayanth.devarayanadurga]

Angle M'OP' = 180 - (90 + theta) = 90 - theta. These angles 90 + theta and 90 - theta add to 180, so they are supplementary angles and sine of supplementary angles are equal.

sin(90 + theta) = sin(180 - (90 + theta))
sin(90 + theta) = sin(90 - theta) = cos theta.

because 90 - theta and theta are complementary angles.

- - - Updated - - -

Angle M'OP' = 180 - (90 + theta) = 90 - theta. These angles 90 + theta and 90 - theta add to 180, so they are supplementary angles and sine of supplementary angles are equal.

sin(90 + theta) = sin(180 - (90 + theta))
sin(90 + theta) = sin(90 - theta) = cos theta.

because 90 - theta and theta are complementary angles.